Approximation - Error Problem, 3

Category: Differential Calculus, Solid Geometry

"Published in Newark, California, USA"

Considering the volume of a spherical shell as an increment of volume of a sphere, find approximately the volume of a spherical shell whose outer diameter is 8 inches and whose thickness is 1/16 inch.

Solution:

To illustrate the problem, it is better to draw the figure as follows

Photo by Math Principles in Everyday Life

If the diameter of a sphere is given in the problem, then the volume of a sphere is

but D = 2R

Since the thickness of a sphere is given and we want to find the volume of a spherical shell, then we have to take the differentials on both sides of the equation. The differential of a volume of a sphere is the same as the volume of a spherical shell. 

Substitute the value of R which is ½ D or 4 in. and dR which is 1/16 in. to the above equation, we have

Therefore, the final answer is

Word Problem - Distance Problem

Category: Algebra

"Published in Suisun City, California, USA"

The fuel consumption for William's car is 30 mi/gal on the highway and 25 mi/gal in the city. On vacation trip of 400 miles, he used 14 gallons of gasoline. How many highway miles and city miles did he drive on this trip?

Solution: 

The given word problem above is about getting the trip distances of highway and city miles that involves the principles of solving two equations, two unkowns because the two fuel consumption for William's car are given. Also, the total vacation trip miles and total gallons of gasoline are given in the problem. 

Let x = be the total trip distance of highway miles
      y = be the total trip distance of city miles
      30 mi/gal = fuel consuption of highway miles
      25 mi/gal = fuel consuption of city miles

If the total vacation trip miles is 400 miles, then we can write the first equation as follows 

If the total gallons of gasoline is 14 gallons, then we can write the second equation as follows

Multiply both sides of the equation by their Least Common Denominator (LCD) which is 150 as follows

The two unknowns of two linear equations can be solved by elimination method. Consider the two linear equations as follows

Multiply the first equation by 5 and -1 at the second equation, we have

                                                    
When you add the two equations, x will be eliminated as follows

Substitute the value of y either of the two equations above as follows

Therefore, the final answers are

        Total Trip Distance of Highway Miles = 300 miles

        Total Trip Distance of City Miles = 100 miles

Differentiation - Rate Problem, 3

Category: Differential Calculus, Physics, Mechanics

"Published in Suisun City, California, USA"

If a ball is thrown vertically upward with a velocity of 80 ft/sec, then its height after t seconds is 

(a) What is the maximum height reached by the ball?
(b) What is the velocity of the ball when it is 96 ft. above the ground on its way up? On its way down?

Solution:

If a ball is thrown in an upward position, then the height of a ball from the ground after t seconds is

To get the velocity of a ball in an upward position, take the derivative of the above equation with respect to time t, as follows

To maximize the height of a ball, equate the above equation to zero or set v = 0, we have

(a) Therefore, the height of a ball is

(b) If a ball is reached to 96 ft. from the ground in an upward position, then the velocity is calculated as follows

Substitute s = 96 ft. to the above equation, we have

Equate each factor to zero and the roots are t = 2 secs and t = 3 secs.

Since the time required to reach the maximum height is 2.5 secs, therefore we have to choose t = 2 secs.

Therefore, the velocity of a ball in an upward direction is

When a ball is reached to its maximum height, then the velocity of a ball in a downward direction at s = 96 ft. from the ground is

Where t = 3 secs (another root of a quadratic equation) after a ball is reached to its maximum height and the ball is about to fall when s = 96 ft. from the ground. 
 

The velocity of a ball is negative because it is a downward direction after it reached to its maximum height. Also, the acceleration which is -16 = ½ g or g = - 32 ft/sec² in the equation is negative because a ball is thrown upward. g is the acceleration due to gravity of an object in earth. g has different values depending in the planets and moon as well.